Question

Solve the following quadratic equation using the quadratic formula: 6x^2 + 3x + 2 = 0

Answer 1

The Quadratic formula is:

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

Given the Quadratic equation:

`6x^2+3x+2=0​`

You can see that it has this form:

`ax^2+bx+c=0`

Then, in this case:

`\begin{gathered} a=6 \\ b=3 \\ c=2 \end{gathered}`

Therefore, you can substitute values into the Quadratic formula:

`x=\frac{-(3)\pm\sqrt[]{(3)^2-4(6)(2)}}{(2)(6)}`

Evaluate:

`\begin{gathered} x=\frac{-3\pm\sqrt[]{^{}-39}}{12} \\ \end{gathered}`

Notice that the Radicand (the value inside the root) is negative.

By definition:

`i=\sqrt[]{-1}`

Then you can rewrite it in this form:

`x=\frac{-3\pm i\sqrt[]{^{}39}}{12}`

Therefore, you get that the answer is:

`\begin{gathered} x_1=\frac{-3-i\sqrt[]{^{}39}}{12} \\ \\ \\ x_2=\frac{-3+i\sqrt[]{^{}39}}{12} \end{gathered}`